Time Series Analysis HW 08 Answer keys - 1 TIME SERIES...

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Unformatted text preview: 1 TIME SERIES ANALYSIS Assignment 8 answer keys 5.5 a)─c) Answer varies. d) Variance of moment estimator: 1+ +4 + 1− + = 1 + 0.6 + 4 0.6 + 0.6 200 1 − 0.6 + 0.6 Variance of innovation estimator: 1/ = 0.005 Variance of MLE: 1 − / = 1 − 0.6 ≈ 0.0237 /200 = 0.0032 e) The MLE is superior over the innovation estimator, which in turn is superior over the moment estimator. 5.9 For the causal AR( ) process the equation (5.2.9) becomes , = 2 !" !# ⋯ ! where, for 1 ≥ + 1, +-, = 3# +, # + ⋯ + 34 +, implies that !, # = 1 for 1 ≥ + 1. So, , Note that, and # # 4 exp (− and 5, *+, − +-, . 1 ) 0, !, # 2 4 # ,/# = 6*+, − +-, . = . The latter *+, − +-, . 1 *!" !# ⋯ !4 # . exp (− 7) + ) *+, − +-, . 90. 2 !, # # = 2 det =4 = det * >4 . = ? @4 =4 # ? 4 = It follows that , 4 ,/48# ,/# det >4 = 4 5" 5# ⋯ 54 4 4 ,/# ,/# # = !" !# ⋯ !4 # , *+, − +-, . *+, − +-, . ? @4 >4 # ? 4 = ) =) . 5, # / !, # = 2 # *det =4 . exp (− + ) *+, − 3# +, ,/48# # 1 7? @4 =4 # ? 4 2 − ⋯ − 34 +, 4. 90. 2 TIME SERIES ANALYSIS 5.10 By Problem 5.9, the least squares estimates 3-# and 3- for a causal zero-mean AR(2) process are the values of 3# and 3 that minimize A 3# , 3 = ? @ = # ? + )*+, − 3# +, where (see large-sample properties of MLE) = # = > # =C That is, A 3# , 3 = 1−3 +# + + ,/B 1−3 −3# 1 + 3 EA 3# , 3 E3 − 3 +, . , −3# 1 + 3 D. 1−3 − 23# 1 + 3 +# + + )*+, − 3# +, ,/B For minimization, we set EA 3# , 3 E3# # = −2 1 + 3 +# + − 2 ) +, # *+, − 3# +, = −23 +# + + Some algebras yield FG 1 − 3# HFG 0 − FG 2 − 3# JFG 1 − ,/B # ,/B +# + + + #+ I − 3 JFG 1 − K − 3 HFG 0 − − 3 +, . . − 3 +, . = 0, − 23# +# + − 2 ) +, *+, − 3# +, +# + + # +# + + + # #+ +# + + + + − 3 +, . = 0. K = 0, # ++ I = 0. For comparison, the corresponding sample Yule−Walker equa5ons are FG 1 − 3# FG 0 − 3 FG 1 = 0, 5.11 FG 2 − 3# FG 1 − 3 FG 0 = 0. For the causal AR(1) process, +L = 3+L # + ML , where NML O~QR 0, F ℎ = 3 |V| , ∀ ℎ. 1−3 and |3| < 1, we have On the other hand, we have +-# = 0, 5" = F 0 (i.e. !" = 1/ 1 − 3 ), and +-, = 3+, 5, # = (i.e. !, # = 1) for 1 ≥ 2. Then, by Problem 5.9, the likelihood of the observations NX# , X O is # and 3 TIME SERIES ANALYSIS 3, = 2 # where Z# = X# and =# = 3, = − ln 2 = − ln 2 Setting E ln we obtain Z#@ =# # Z# + X − 3X# [, 1 1 1 − ln − X# 1 − 3 + X − 3X# 2 1−3 2 1 1 − ln + ln 1 − 3 − X# − 23X# X + X . 2 2 3, 3, E3 =− 3- = 5.12 1 2 − ln E ln E exp Y− F 0 = 1/ 1 − 3 . So, the log-likelihood is ln # det =# 1 =− + 2X# X X# + X For the causal AR(1) process, +L = 3+L 2 3 1−3 1 X# X = 0, X# − 23X# X + X and # + G = X# − X . 2 X# + X + ML , where NML O~QR 0, F ℎ = = 0, 3 |V| , ∀ ℎ. 1−3 and |3| < 1, we have On the other hand, we have +-# = 0, 5" = F 0 (i.e. !" = 1/ 1 − 3 ), and +-, = 3+, 5, # = (i.e. !, # = 1) for 1 ≥ 2. Then, the MLE 3- is the value of 3 that minimizes 1 1 _ 3 = ln J A 3 K + ) ln !, ,/# where, by Problem 5.9, # 1 1 = ln J A 3 K + ln A 3 = ?#@ =# # ?# + )*+, − 3+, # . = +# 1 − 3 ,/ Setting = FG 0 − 2 FG 1 3 + + )*+, − 3+, # . ,/ FG 0 − +# + + E_ 3 1 EA 3 = + E3 A 3 E3 23 1−3 1 , 1−3 3 . # and 4 TIME SERIES ANALYSIS to zero, we obtain 1−3 where Some algebras yield − 1 HFG 0 − EA 3 + 23A 3 = 0, E3 EA 3 = −2 FG 1 + 2 FG 0 − +# + + E3 +# + + I 3B − − 2 FG 1 3 − 3. + 1 FG 0 − +# + + 3 + FG 1 = 0. ...
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